Let $(X,d)$ be a Polish space.
Suppose that $(\mu_n)$ is a sequence of probability measures on $X$ such that $\mu_n\to \mu$ weakly, where $\mu$ is a probability measure on $X$.
Is it true that there exists a $\sigma$-additive function $\nu:\mathcal{B}(X)\to\mathbb{R}$ such that
$$
\mu_n(A)\leq \nu(A)\quad\text{ for all }n,\:A\in\mathcal{B}(X)
$$
?
If not, which conditions can be assumed on $X$ so that the assertion is true? $X$ compact for instance?
Not possible to find $\nu$ even in nice situations. For example $\delta_{\frac 1 n} \to \delta_0$ weakly on $[0,1]$ but $\delta_{\frac 1 n} \leq \nu$ for all $n$ implies $\nu [0,1]=\infty$ because $\nu \{\frac 1 n\}\geq 1$ for all $n$. Hence we cannot have a $\nu$ which takes only finite values.